Tag: maths

Plane ${P} _{1}$ is parallel to $\overrightarrow{a}$ and $\overrightarrow{b}$.
The normal to ${P} _{1}$ is along $\overrightarrow{a}\times \overrightarrow{b}$.
Plane ${P} _{2}$ is parallel to $\overrightarrow{c}$ and $\overrightarrow{d}$.
The normal to ${P} _{2}$ is along $\overrightarrow{c}\times \overrightarrow{d}$.
$\overrightarrow{A}$ is along the line of intersection of planes ${P} _{1}$ and ${P} _{2}$.
$\therefore \overrightarrow{A}$ is along $\left(\overrightarrow{a}\times\overrightarrow{b}\right)\times\left(\overrightarrow{c}\times\overrightarrow{d}\right)$
$\overrightarrow{a}\times\overrightarrow{b}=\left|\begin{matrix} \hat{i} &\hat{j}  &\hat{k}  \ 0 & 2 & 3 \ 0 &4  &-3  \end{matrix}\right|$
$=\left(-6-12\right)\hat{i}-0.\hat{j}+0.\hat{k}$ on simplification
$=-18\hat{i}$
$\left(\overrightarrow{a}\times\overrightarrow{b}\right)\times\left(\overrightarrow{c}\times\overrightarrow{d}\right)$
$\overrightarrow{c}\times\overrightarrow{d}=\left|\begin{matrix} \hat{i} &\hat{j}  &\hat{k}  \ 0 & 1 & -1 \ 3 &3  &0 \end{matrix}\right|$
$=\left(0+3\right)\hat{i}-\left(0+3\right)\hat{j}+\left(0+3\right)\hat{k}$ on simplification
$=3\hat{i}-3\hat{j}-3\hat{k}$
$=3\left(\hat{i}-\hat{j}-\hat{k}\right)$
The angle between $\overrightarrow{A}$ and $2\hat{i}+\hat{j}-2\hat{k}$ is  $\theta$
$\cos{\theta}=\dfrac{\overrightarrow{A}}{\left|\overrightarrow{A}\right|}.\dfrac{\left(2\hat{i}+\hat{j}-2\hat{k}\right)}{3\sqrt{2}}$
$\pm\dfrac{\left(\hat{j}-\hat{k}\right).\left(2\hat{i}+\hat{j}-2\hat{k}\right)}{3\sqrt{2}}$
$\pm\dfrac{1}{\sqrt{2}}$
and $\cos{\theta}=\pm\dfrac{1}{\sqrt{2}}$
$\Rightarrow \theta=\dfrac{\pi}{4},\dfrac{3\pi}{4}$

Plane ${p} _{1}$ is parallel to $\overrightarrow{a}$ and $\overrightarrow{b}$ the normal to ${p} _{1}$ is along $\overrightarrow{a}\times \overrightarrow{b}$ 
Plane ${p} _{2}$ is parallel to $\overrightarrow{c}$ and $\overrightarrow{d}$ the normal to ${p} _{2}$ is along $\overrightarrow{c}\times \overrightarrow{d}$
$\overrightarrow{A}$ is along the line of intersection of planes ${p} _{1}$ and ${p} _{2}$ 
$\therefore \overrightarrow{A}$ is along $\left(\overrightarrow{a}\times \overrightarrow{b}\right)\times \left(\overrightarrow{c}\times \overrightarrow{d}\right)$
$\overrightarrow{a}\times \overrightarrow{b}=\left[\begin{matrix} \hat{i} & \hat{j} & \hat{k} \ 0 & 2 &  3\ 0 & 4 & -3 \end{matrix}\right]$
$=\hat{i}\left(-6-12\right)-\hat{j}\left(0-0\right)+\hat{k}\left(0\right)$
$=-18\hat{i}$
$\overrightarrow{c}\times \overrightarrow{d}=\left|\begin{matrix} \hat{i} & \hat{j} & \hat{k} \ 0  & 1 & -1 \ 3 & 3 & 0 \end{matrix}\right|$
$=\hat{i}\left(0+3\right)-\hat{j}\left(0+3\right)+\hat{k}\left(0-3\right)$
$=3\hat{i}-3\hat{j}-3\hat{k}$
$=3\left(\hat{i}-\hat{j}-\hat{k}\right)$
$ \therefore \overrightarrow{A}$ is along $\hat{i}\times \left(\hat{i}-\hat{j}-\hat{k}\right)=\hat{j}-\hat{k}$
The angle between $\overrightarrow{A}$ and $2\hat{i}+\hat{j}-2\hat{k}$ is $\theta$
$\cos{\theta}=\dfrac{\overrightarrow{A}}{\left|\overrightarrow{A}\right|}.\dfrac{\left(2\hat{i}+\hat{j}-2\hat{k}\right)}{3}$
$   =\pm \dfrac{\left(\hat{j}-\hat{k}\right)\left(2\hat{i}+\hat{j}-2\hat{k}\right)}{3\sqrt{2}}$
$=\pm\dfrac{\left(1+2\right)}{3\sqrt{2}} = \pm \dfrac{1}{\sqrt{2}}$
and $\cos{\theta}=\pm \dfrac{1}{\sqrt{2}}$
$\Rightarrow \theta=\dfrac{\pi}{4},\dfrac{3\pi}{4}$

The plane ${p} _{1}$ contains the vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ into normal is along $\overrightarrow{a}\times \overrightarrow{b}$
The normal to plane ${p} _{2}$ is along  $\overrightarrow{c}\times \overrightarrow{d}$.
$\left(\overrightarrow{a}\times \overrightarrow{b}\right)\times \left(\overrightarrow{c}\times \overrightarrow{d}\right)=0$
$\Rightarrow$ two normals are parallel
$\therefore$ the angle between the planes is zero

Plane1 :$3x+4y-5z+1=0$

The equations of the plane which is equally inclined to the planes $x-y+z-3=0$ and $x+y+z+4=0$ is/are- 
$\dfrac { x-y+z-3 }{ \sqrt { { 1 }^{ 2 }+{ 1 }^{ 2 }+{ 1 }^{ 2 } }  } \pm \dfrac { x+y+z+4 }{ \sqrt { { 1 }^{ 2 }+{ 1 }^{ 2 }+{ 1 }^{ 2 } }  } =0$
$\Rightarrow x+z=-1$ and $y=\dfrac { -7 }{ 2 } $
If the plane contains origin
Then desired planes are $x+z=0$ & $y=0$

Ans: A,D

$cos \theta = \dfrac{ a _1 \, a _2 + b _1 \, b _2 + c _1 \, c _2}{\sqrt{a _1^2 + b _1^2 + c _1^2} \sqrt{a _2^2 + b^2 _2 + c _2^2}}$

Equation of the planes is $2x-4y+2z+3=0$ and $3x-2y+6z+8=0$
Then equation of the plane bisection the angles between them are
$\displaystyle \frac { 2x-4y+2z+3 }{ \sqrt { 4+16+4+9 }  } =\pm \frac { 3x-2y+6z+8 }{ \sqrt { 9+4+36+64 }  } $
$\displaystyle \Rightarrow \frac { 2x-4y+2z+3 }{ \sqrt { 33 }  } =\pm \frac { 3x-2y+6z+8 }{ \sqrt { 113 }  } $
$\Rightarrow 5x-y-4z-22=0$ and $23x-13y+32z+26=0$

The angle between two intersecting planes is equal to the acute angle determined by the normal vectors of the two planes.

I am using the constant $\lambda$ instead of $k$ to avoid the confusion. 

Consider $P : bx+cz+d=0$
a) Equation of $YOZ$ plane is $x=0$
Since, $(i).(bj+ck)=0$
Therefore, $P$ is perpendicular to $YOZ$

b)  Equation of $ZOX$ plane is $y=0$
Since, $(j).(bj+ck)=b \neq 0$
Therefore, $P$ is not perpendicular to $ZOX$

c)  Equation of $XOY$ plane is $z=0$
Since, $(k).(bj+ck)=c \neq 0$
Therefore, $P$ is not perpendicular to $XOY$

d)  Consider. $z=k$
Since, $(k).(bj+ck)=c \neq 0$
Therefore, $P$ is not perpendicular to $z=k$

Ans: A